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  leetcode-鸡蛋掉落
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  <h1 id="鸡蛋掉落-难度：困难"><a href="#鸡蛋掉落-难度：困难" class="headerlink" title="鸡蛋掉落(难度：困难)"></a>鸡蛋掉落(难度：困难)</h1><p>你将获得 K 个鸡蛋，并可以使用一栋从 1 到 N  共有 N 层楼的建筑。</p>
<p>每个蛋的功能都是一样的，如果一个蛋碎了，你就不能再使用它，否则可以继续丢，鸡蛋的性能不会随着丢的次数增加而有所损耗。</p>
<p>假设存在一个中间楼层F，满足 0 &lt;= F &lt;= N 任何从高于 F 的楼层落下的鸡蛋都会碎，从 F 楼层或比它低的楼层落下的鸡蛋都不会破。</p>
<p>每次移动，你可以取一个鸡蛋（如果你有完整的鸡蛋）并把它从任一楼层 X 扔下（满足 1 &lt;= X &lt;= N）。</p>
<p>你的目标是确切地知道 F 的值是多少。</p>
<p>无论 F 的初始值如何，你确定 F 的值的最小移动次数是多少？</p>
<p>示例 1：</p>
<blockquote>
<p>输入：K = 1, N = 2<br>输出：2<br>解释：<br>鸡蛋从 1 楼掉落。如果它碎了，我们肯定知道 F = 0 。<br>否则，鸡蛋从 2 楼掉落。如果它碎了，我们肯定知道 F = 1 。<br>如果它没碎，那么我们肯定知道 F = 2 。<br>因此，在最坏的情况下我们需要移动 2 次以确定 F 是多少。</p>
</blockquote>
<p>示例 2：</p>
<blockquote>
<p>输入：K = 2, N = 6<br>输出：3</p>
</blockquote>
<p>示例 3：</p>
<blockquote>
<p>输入：K = 3, N = 14<br>输出：4</p>
</blockquote>
<a id="more"></a>
<h2 id="动态规划"><a href="#动态规划" class="headerlink" title="动态规划"></a>动态规划</h2><p>这个方法可以观看李永乐的视频，比较好理解</p>
<p><a href="https://www.youtube.com/watch?v=mLV_vOet0ss" target="_blank" rel="noopener">复工复产找工作？先来看看这道面试题：双蛋问题</a></p>
<p>维护一个状态数组dp,<code>dp[i][j]</code>代表一共有 i 层楼的情况下，使用 j 个鸡蛋的最少实验的次数。</p>
<p>说明：</p>
<p>i 表示的是楼层的大小相对，不是绝对高度（第几层）的意思，例如楼层区间 [8, 9, 10] 的大小为 3。<br>j 表示可以使用的鸡蛋的个数，它是约束条件。</p>
<p><img src="https://img-blog.csdnimg.cn/20200411111010534.png" alt=""><br>对于这个数组，我们可以对它的一些特殊情况进行初始化</p>
<blockquote>
<p>只有一个鸡蛋：最少移动数就是楼层数。只有一层楼：永远只需要移动一次</p>
</blockquote>
<p><img src="https://img-blog.csdnimg.cn/20200411111116868.png" alt="在这里插入图片描述"><br>那么接下来对于每一个高度<code>i</code>我们就需要进行枚举，在0到<code>i</code>之间选一个楼层k开始试验，k从1开始直到<code>i</code>，就可以依次填写这个数组</p>
<p>对于当前的楼层k，k &gt;= 1 且 k &lt;= i：</p>
<p>如果鸡蛋破碎，实验就得在 k 层以下做（不包括 k 层），这里已经使用了一个鸡蛋，简单来说就是楼层数减一，鸡蛋数减一，即求：<code>dp[k - 1][j - 1]</code>；</p>
<p>如果鸡蛋完好，那么就要去k楼层以上进行试验。那么对于表中的<code>i</code>层楼，剩下的楼层数就是<code>i- k</code>，所以这个时候即求：<code>dp[i - k][j])</code></p>
<p>那么对于同一个k取值时存在摔碎与没摔碎两种情况，我们要去求最差情况，也就是两者之间的最大值</p>
<p>对于不同k之间的取值，我们需要取两者之间的最小值</p>
<p>代码如下：</p>
<figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">var</span> superEggDrop = <span class="function"><span class="keyword">function</span>(<span class="params">K, N</span>) </span>&#123;</span><br><span class="line">    <span class="keyword">let</span> dp = <span class="keyword">new</span> <span class="built_in">Array</span>(N + <span class="number">1</span>)</span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">let</span> i = <span class="number">0</span>; i &lt;= N; i++)&#123;</span><br><span class="line">        dp[i] = <span class="keyword">new</span> <span class="built_in">Array</span>(K + <span class="number">1</span>).fill(i)</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">for</span>(<span class="keyword">let</span> i = <span class="number">2</span>; i &lt;= N; i++)&#123;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">let</span> j = <span class="number">2</span>; j &lt;= K; j++)&#123;</span><br><span class="line">            <span class="keyword">for</span> (<span class="keyword">let</span> k = <span class="number">1</span>; k &lt;= i; k++) &#123;</span><br><span class="line">                dp[i][j] = <span class="built_in">Math</span>.min(dp[i][j], <span class="built_in">Math</span>.max(dp[k - <span class="number">1</span>][j - <span class="number">1</span>], dp[i - k][j]) + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> dp[N][K]</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
<p>在上述解法中，由于开辟了一个二维数组dp，所以空间复杂度为O(NK)。填表本身的时间复杂度为O(NK)， 但是对于表中的每一项（也就是每一个高度）,都要对从哪里开始实验进行枚举，所以时间复杂度变为O(KN^2)</p>
<p>所以不出意外，这道题提交直接TLE</p>
<h2 id="动态规划（重写状态转移方程）"><a href="#动态规划（重写状态转移方程）" class="headerlink" title="动态规划（重写状态转移方程）"></a>动态规划（重写状态转移方程）</h2><p>这道题可以进行一下逆行思维，将状态转移方程<code>dp[i][j]</code>看成，当你有j和鸡蛋，最多可以走i步，你最多可以验证<code>dp[i][j]</code>高的楼层</p>
<p>那么对于<code>dp[i][j]</code>来说，<code>dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] + 1</code></p>
<figure class="highlight plain"><figcaption><span>j 个蛋，且只能操作 i 次了，所能确定的楼层。</span></figcaption><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line">&#96;&#96;&#96;dp[i - 1][j]&#96;&#96;&#96;：蛋没碎，因此该部分决定了所操作楼层的上面所能容纳的楼层最大值</span><br><span class="line">&#96;&#96;&#96;dp[i-1][j-1]&#96;&#96;&#96;：蛋碎了，因此该部分决定了所操作楼层的下面所能容纳的楼层最大值</span><br><span class="line"></span><br><span class="line">那么最终dp数组中最后一列中刚刚大于楼层数N的那一个数所在的行数就是最小操作数</span><br><span class="line"></span><br><span class="line">![](https:&#x2F;&#x2F;img-blog.csdnimg.cn&#x2F;20200411115431149.png)</span><br><span class="line">代码如下：</span><br><span class="line"></span><br><span class="line">&#96;&#96;&#96;javascript</span><br><span class="line">var superEggDrop &#x3D; function(K, N) &#123;</span><br><span class="line">    let dp &#x3D; new Array(N)</span><br><span class="line">    for(let i &#x3D; 0; i &lt; N; i++)&#123;</span><br><span class="line">        if(i &#x3D;&#x3D;&#x3D; 0)&#123;</span><br><span class="line">            dp[i] &#x3D; new Array(K).fill(1)</span><br><span class="line">        &#125;</span><br><span class="line">        else&#123;</span><br><span class="line">            dp[i] &#x3D; new Array(K).fill(i + 1)</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    for(let i &#x3D; 1; i &lt; N; i++)&#123;</span><br><span class="line">        for(let j &#x3D; 1; j &lt; K; j++)&#123;</span><br><span class="line">            dp[i][j] &#x3D; dp[i - 1][j] + dp[i - 1][j - 1] + 1</span><br><span class="line">        &#125;</span><br><span class="line">        if(dp[i][K - 1] &gt;&#x3D; N)&#123;</span><br><span class="line">            return i + 1</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    return dp[0][K - 1]</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
<h2 id="再优化"><a href="#再优化" class="headerlink" title="再优化"></a>再优化</h2><p>对于上述的dp数组我们发现，第 f 次操作结果只和第 f-1 次操作结果相关，因此我们可以把dp压缩成为一个一维数组，我们发现纵向最多允许操作数是顺序增长的，所以我们可以把纵向最大操作数取消，使用一个变量count递增来代替，保留鸡蛋数，<code>dp[i]</code>代表有i个鸡蛋时，当前count次操作后可以确认的楼层数。.</p>
<p>代码如下：</p>
<figure class="highlight javascript"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">var</span> superEggDrop = <span class="function"><span class="keyword">function</span>(<span class="params">K, N</span>) </span>&#123;</span><br><span class="line">    <span class="keyword">let</span> dp = <span class="keyword">new</span> <span class="built_in">Array</span>(K + <span class="number">1</span>).fill(<span class="number">0</span>)</span><br><span class="line">    <span class="keyword">let</span> count = <span class="number">0</span></span><br><span class="line">    <span class="keyword">while</span>(dp[K] &lt; N)&#123;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">let</span> i = K; i &gt; <span class="number">0</span>; i--)&#123;</span><br><span class="line">            dp[i] = dp[i] + dp[i - <span class="number">1</span>] + <span class="number">1</span></span><br><span class="line">        &#125;</span><br><span class="line">        count ++</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> count</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>
 
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<link rel="stylesheet" href="/css/clipboard.css">

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      copyHtml += '<i class="ri-file-copy-2-line"></i><span>COPY</span>';
      copyHtml += '</button>';
      $(".highlight .code pre").before(copyHtml);
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        target: function(trigger) {
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        $icon.removeClass('ri-file-copy-2-line');
        $icon.addClass('ri-checkbox-circle-line');
        let $span = $($btn.find('span'));
        $span[0].innerText = 'COPIED';
        
        wait(function () { // 等待两秒钟后恢复
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          $span[0].innerText = 'COPY';
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        $span[0].innerText = 'COPY FAILED';
        
        wait(function () { // 等待两秒钟后恢复
          $icon.removeClass('ri-time-line');
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        }, 2000);
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</script>


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